推 ONIspirit:感謝~~~ 59.104.235.59 02/25 00:41
※ 引述《ONIspirit (嘎)》之銘言：
: find the arc length of the curve y = ln(1-x^2)
: from x = 0 to x = 1/4
dy/dx=(-2x)/(1-x^2)
let a=0 b=1/4
b
所求=∫√[1+(dy/dx)^2] dx
a
=∫√[1-2x^2+x^4+4x^2/(1-x^2)^2] dx
=∫√[(1+x^2)^2/(1-x^2)^2] dx
=∫(1+x^2)/(1-x^2) dx
=∫-1+[2/(1+x)(1-x)] dx
b b b
=-x│-ln(1-x)│+ln(1+x)│
a a a
=ln(5/3)-1/4
參考參考...
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